Question: What is the extraneous solution to these equations? $\dfrac{x^2 + 16}{x + 5} = \dfrac{-15x - 34}{x + 5}$
Solution: Multiply both sides by $x + 5$ $ \dfrac{x^2 + 16}{x + 5} (x + 5) = \dfrac{-15x - 34}{x + 5} (x + 5)$ $ x^2 + 16 = -15x - 34$ Subtract $-15x - 34$ from both sides: $ x^2 + 16 - (-15x - 34) = -15x - 34 - (-15x - 34)$ $ x^2 + 16 + 15x + 34 = 0$ $ x^2 + 50 + 15x = 0$ Factor the expression: $ (x + 5)(x + 10) = 0$ Therefore $x = -5$ or $x = -10$ At $x = -5$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -5$, it is an extraneous solution.